- #1

transgalactic

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if f(x) is differentiable on x_0

prove that

http://img102.imageshack.us/img102/3189/51290270sj1.th.gif

??

[itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/itex]

now let [itex] u = x+h \Rightarrow x = u-h [/itex]

So [itex] f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} [/itex]

Since [itex] h \to 0, f'(u-h) \to f'(u) [/itex]

Therefore [itex] f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} [/itex]

[itex] f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)[/itex]

[itex] 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} [/itex]

Hence, [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} [/itex]

i understand this solution but

how to pick those variables??

and why it didnt use the point x_0

that they presented

prove that

http://img102.imageshack.us/img102/3189/51290270sj1.th.gif

??

[itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/itex]

now let [itex] u = x+h \Rightarrow x = u-h [/itex]

So [itex] f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} [/itex]

Since [itex] h \to 0, f'(u-h) \to f'(u) [/itex]

Therefore [itex] f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} [/itex]

[itex] f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)[/itex]

[itex] 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} [/itex]

Hence, [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} [/itex]

i understand this solution but

how to pick those variables??

and why it didnt use the point x_0

that they presented

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